# Math Help - Fibonacci Look-a-like

1. ## Fibonacci Look-a-like

Define a fibonacci series generated by {a,b} such as,
$u_1=a$
$u_2=b$
$u_{n+2}=u_{n+1}+u_{n}$
(The original fibonacci series is generated by {1,1})
( $a\not =0$)

Then there are a number of interesting properties:
1)If $K$ is the finite continued fraction for $b/a$
then $\frac{u_{n+1}}{u_n}=[1;1,1...,1,K]$
Where the $1$ appears $n-2$ times.

2)Thus, from here we have that $\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\psi$
(Thus, any fibonacci series generated by any two real numbers converges to the divine proportion).

3)The formula for $u_n$ is given by
$u_n=F(n-2)a+F(n-1)b$
where $F(n)$ is the n-th fibonacci number.
But by Binet's formula we have that,
for the n-th fibonacci number we can find a formula for $u_n$ but it is rather messy and will be omitted. Giving a second method for proving statement 2.

2. I think you mean $\phi$ and not $\psi$. Very interesting, though.

3. Originally Posted by Treadstone 71
I think you mean $\phi$ and not $\psi$. Very interesting, though.

$
\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi
$

4. Originally Posted by Natasha1
$
\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi
$
That is not important

5. The proof of the second statement is rather neat.

$2)\;\;\lim_{n\to\infty}\frac{u_{n+1}}{u_n}\:=\: \phi$

The limit of the ratio of consecutive Fibonacci numbers is already established.

. . . . . $[1]\;\lim_{n\to\infty}\frac{F_{n+1}}{F_n}\:=\:\phi \qquad\qquad[2]\;\lim_{n\to\infty}\frac{F_n}{F_{n+1}} \:=\:\frac{1}{\phi}$

We have: $u_n\:=\:a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)$

The ratio is: $R\;=\;\frac{u_{n+1}}{u_n}\,=\,\frac{a\!\cdot\!F(n-1) + b\!\cdot\!F(n)}{a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)}$

Divide top and bottom by $F_{n-1}:\;\;R\;=\;\frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}$

Take the limit: $\lim_{n\to\infty}R\;=\;\lim_{n\to\infty}\left( \frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}\right)$

From [1] amd [2], this becomes: $\frac{a + b\!\cdot\!\phi}{a\!\cdot\!\frac{1}{\phi} + b}$

Multiply top and bottom by $\phi:\;\;\frac{\phi(a + b\!\cdot\!\phi)}{a + b\!\cdot\!\phi}\;=\;\phi$

6. Originally Posted by Soroban
The proof of the second statement is rather neat.

[size=3]
My proof is simpler.
Because of condition 1), since this countinued fractions converges to $[1:1,1,...]=\phi$.

7. Hello, ThePerfectHacker!

My proof is simpler.
Of course it is . . . I wasn't claiming otherwise.

I was demonstrating a straight algebraic approach to the limit
. . for those not familiar with continued fractions.

Please understand: I never ever compete with other posters here.
I may post an alternate approach or even an improvement
. . . but never with a snicker or a sneer, implied or otherwise.